Where does the 2! come from in Taylor series?

[LostInSpace]

I am a bit confused about taylor approximation. Taylor around $$x_0$$ yields
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + O(x^2)$$

which is the tangent of f in $$x_0$$, where
$$f'(x) = f'(x_0) + f''(x_0)(x-x_0) + O(x^2)$$

which adds up to
$$f(x) &=& f(x_0) + (f'(x_0) + f''(x_0)(x-x_0) + O(x^2))(x-x_0)+O(x^2) \\ &=& f(x_0) + f'(x_0)(x-x_0) + f''(x_0)(x-x_0)^2 + O(x^3)$$
But it should be
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + O(x^3)$$

Where does the 2! come from? Is this approach completely incorrect?

 

[arildno]

You have ignored in line 3 the O(x^(2))-term from the expansion of f(x).
Hence line 3 is not accurate to O(x^(3)), it's only accurate to O(x^(2)).

 

[Hurkyl]

Do you remember how to derive, from the limit definition of the derivative, the differential approximation formula:

$$f(x+\epsilon ) = f(x) + \epsilon f'(x) + \epsilon \delta(x, \epsilon)$$

Where $\lim_{\epsilon \rightarrow 0} \delta(x, \epsilon) = 0$?

Try writing the second derivative with limits, and see if any approach suggests itself.